Answer:
4.35 * 10^-8 M
Explanation:
Since the concentration of the hydronium ion= 2.3 X 10^-7 M
And we know that;
[H3O^+] [OH^-] = 1 * 10^-14
[H3O^+] = concentration of the hydronium ion
[OH^-] = concentration of the hydroxide ion
So;
[OH^-] =1 * 10^-14/[ H3O^+]
But [H3O^+] = 2.3 X 10^-7 M
[OH^-] = 1 * 10^-14/2.3 X 10^-7
[OH^-] = 4.35 * 10^-8 M
<span>C4H10 + 6.5 O2 ----> 4CO2 + 5H2O
2C4H10 + 13 O2 ----> 8CO2 + 10H2O
1. Count the C on the left (4), put a 4 where the C on the right.
2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O
3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you need 6.5 on the left.
4. Now multiply everything on the equation by two so you have nice integer numbers.
5. check you have the same amount of everything on each side.
Example C: left 8, right 8, etc.
I hope this helps. :)</span><span>
</span>
Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2
From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams
Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams
From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid
To get the amount of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
</span>amount of Al = (<span>5.890 x 34) / 294 = 0.6811 grams</span>
Answer:
13
Explanation:
proton plus neutron equal to atomic mass
His measurements are precise since his pH values are close to each other in a way that it was repeated in all measurements. On the contrary to accuracy, it is the closeness to the actual pH value he should have achieved. Therefore, Jose's results are precise but not accurate since his value is not close to the actual value of pH 4.
